Answer
$\frac{4x-7}{x^2-x+1}$
Work Step by Step
Step 1. Factor the third denominator, we have $x^3+1=(x+1)(x^2-x+1)$
Step 2. The least common denominator can be identified as $(x+1)(x^2-x+1)$
Step 3. Convert each expression to have the least common denominator then perform operations:
$\frac{4}{x+1}+\frac{1}{x^2-x+1}-\frac{12}{x^3+1}=\frac{4(x^2-x+1)}{(x+1)(x^2-x+1)}+\frac{x+1}{(x+1)(x^2-x+1)}-\frac{12}{(x+1)(x^2-x+1)}=\frac{4x^2-3x-7}{(x+1)(x^2-x+1)}=\frac{(x+1)(4x-7)}{(x+1)(x^2-x+1)}=\frac{4x-7}{x^2-x+1}$ where $x\ne -1, x^2-x+1\ne 0$
